Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $r = \dfrac{n - 8}{4n^2 - 4n - 360} \div \dfrac{-3n^2 + 30n}{2n^3 - 2n^2 - 180n} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $r = \dfrac{n - 8}{4n^2 - 4n - 360} \times \dfrac{2n^3 - 2n^2 - 180n}{-3n^2 + 30n} $ First factor out any common factors. $r = \dfrac{n - 8}{4(n^2 - n - 90)} \times \dfrac{2n(n^2 - n - 90)}{-3n(n - 10)} $ Then factor the quadratic expressions. $r = \dfrac {n - 8} {4(n + 9)(n - 10)} \times \dfrac {2n(n + 9)(n - 10)} {-3n(n - 10)} $ Then multiply the two numerators and multiply the two denominators. $r = \dfrac {(n - 8) \times 2n(n + 9)(n - 10) } { 4(n + 9)(n - 10) \times -3n(n - 10)} $ $r = \dfrac {2n(n + 9)(n - 10)(n - 8)} {-12n(n + 9)(n - 10)(n - 10)} $ Notice that $(n + 9)$ and $(n - 10)$ appear in both the numerator and denominator so we can cancel them. $r = \dfrac {2n\cancel{(n + 9)}(n - 10)(n - 8)} {-12n\cancel{(n + 9)}(n - 10)(n - 10)} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $r = \dfrac {2n\cancel{(n + 9)}\cancel{(n - 10)}(n - 8)} {-12n\cancel{(n + 9)}(n - 10)\cancel{(n - 10)}} $ We are dividing by $n - 10$ , so $n - 10 \neq 0$ Therefore, $n \neq 10$ $r = \dfrac {2n(n - 8)} {-12n(n - 10)} $ $ r = \dfrac{-(n - 8)}{6(n - 10)}; n \neq -9; n \neq 10 $